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1. Irreducible Scheme a Topological Space X Is Irreducible If X Is Nonempty and Can Not Be a Union of Two Proper Closed Subsets

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1. Irreducible Scheme a Topological Space X Is Irreducible If X Is Nonempty and Can Not Be a Union of Two Proper Closed Subsets 1. Irreducible Scheme A topological space X is irreducible if X is nonempty and can not be a union of two proper closed subsets. A nonempty closed subset Z of X is irreducible if Z equipped with the induced topology from X is an irreducible topological space. A point ξ of an irreducible closed subset Z of X is called a generic point if Z = fξg: Proposition 1.1. A topological space X is irreducible if and only if the intersection of any two nonempty open subsets of X is nonempty. Proof. Suppose that X is irreducible. Assume that there exist two nonempty open subsets U1;U2 so that U1 \ U2 = ;: Denote Zi = X n Ui for i = 1; 2: Then Z1 and Z2 are proper closed subsets of X: Since U1 \ U2 = ;;Z1 [ Z2 = X: This leads to a contradiction that X is irreducible. Hence any two nonempty open subsets of X is nonempty. Conversely, suppose that the intersection of any two nonempty open subsets of X is nonempty. Assume that X = Z1 [ Z2 is a union of closed subsets of X: Denote Ui = X n Zi for i = 1; 2: Then U1 \ U2 = ;: This shows that either U1 = ; or U2 = ;: This is equivalent to say that Z1 = X or Z2 = X: Hence X is irreducible. Definition 1.1. A scheme is irreducible if its underlying topological space is irreducible. Lemma 1.1. Let X = Spec A be the spectrum of a ring A and D(f) be the distinguished open set associated with f 2 A: Then D(f) = ; if and only if f is nilpotent. Proof. Let f be nilpotent. Then f n = 0 for some n > 0: In other words, f n 2 p for all p 2 X: Since p is prime, f 2 p: We see that f 2 p for all p 2 X: Hence p 2 V (f) for all p 2 X: We obtain X = V (f): Hence D(f) = ;: Conversely, suppose D(f) = ;: Then X = V (f): This shows that f 2 p for all p: Hence T f 2 p2X p = Nil(A): (The nilradical of A:) Hence f is nilpotent. Proposition 1.2. Let X = Spec A be the spectrum of a ring A. Then X is irreducible if and only if its nil radical Nil(A) is a prime ideal. Proof. Let us assume that ξ = Nil(A) is a prime ideal. Then the Nil(A) is the intersection of all prime ideals of A: Hence p 2 V (ξ) for all p 2 X: Hence X = V (ξ): Notice that V (ξ) is a closed subset of X containing ξ: Then V (ξ) contains the closure of ξ: Moreover, if V (I) is a closed subset of X containing ξ; then ξ ⊃ I: Thus V (ξ) ⊂ V (I): This implies that V (ξ) is in fact the closure of ξ: In other words, ξ is the generic point. Hence all nonempty open subsets of X contain ξ: Hence the intersection of any two nonempty open subsets of X contains ξ: Suppose that X is irreducible. Let fg 2 Nil(A): To show that Nil(A) is a prime, we need to show that either f 2 Nil(A) or g 2 Nil(A): Then D(fg) = D(f)\D(g): If both D(f) and D(g) are nonempty, then f and g are not nilpotent. Since X is irreducible and D(f) and D(g) are nonempty open subsets of X; D(f) \ D(g) = D(fg) is nonempty. This implies that fg is not nilpotent, i.e. fg 62 Nil(A): This leads to the contradiction to the assumption that fg 2 Nil(A): Thus either D(f) = ; or D(g) = ;: Hence either f or g is nilpotent, i.e. f 2 Nil(A) or g 2 Nil(A): Proposition 1.3. Let X = Spec A: For p 2 Spec A; V (p) is an irreducible closed subset. Conversely, every irreducible closed subset of X is of the form V (p) for some p 2 X: 1 2 Proof. Let Z = V (I) be a nonempty closed subset of X where I is a radical ideal of X: Suppose I is not prime. Then we can choose f; g 2 A so that fg 2 I but f; g 62 I: In this case, we consider (f) + I and (g) + I: Then V ((f) + I) and V ((g) + I) are proper closed subsets of V (I): Moreover, V ((f) + I) [ V ((g) + I) = V ((fg) + I) = V (I) = Z: This implies that Z is not irreducible. If Z is not irreducible, Z is a union of two proper closed subsets V (I1) and V (I2): Assume that I1 and I2 are radical ideals of A: Then I1 ⊃ I and I2 ⊃ I: Choose fi 2 Ii n I: Then (fi) + I ⊃ I for i = 1; 2: Then V ((fi) + I) ⊃ V (Ii): We see that V ((f1f2) + I) = V ((f1) + I) [ V ((f2) + I) = V (I): p p Then (f1f2) + I = I: Since (f1f2) ⊂ (f1f2) + I ⊂ (f1f2) + I; we see that f1f2 2 I: This shows that I is not a prime. If p is a prime ideal of a ring A; then V (p) is the closure of p in Spec A: Proposition 1.4. Let X be a scheme. Suppose X is irreducible. Then X has a unique generic point. Proof. Suppose X = Spec A: We have seen that ξ = Nil(A) is the unique generic point. Since X is irreducible, any affine open subset U of X is also irreducible. Suppose X (not necessarily affine) has two generic points ξ and η: Then ξ and η are also generic points of U of X: Since U is affine, ξ = η by the uniqueness of the generic point of an affine scheme. Hence we proved the uniqueness of generic point of a scheme. Now let us prove the existence. The affine open subsets of X forms a basis for the Zariski topology of X: Since X is irreducible, the intersection of any two nonempty affine open subsets of X are nonempty. Hence there exists a nonempty affine open set contained in the intersection of any two nonempty affine open subsets of X: Let U; V be nonempty affine open sets. Choose ξ and η so that ξ and η are generic point of U and V respectively. Choose W nonempty affine open so that W ⊂ U \ V: Then we know both ξ and η are generic point of W: By uniqueness of the generic point of W; ξ = η: This proves that all the affine open subsets of X share the same generic point ξ. Now let us prove that ξ is the generic point of X: Let x be any point of X and U be any open neighborhood of x: Since X is a scheme, we can choose an affine open neighborhood V of x contained in U: Since ξ is a generic point of V; U contains ξ: Hence x lies in the closure of fξg: Thus X = fξg: Proposition 1.5. Let X be a scheme. Any irreducible closed subset of X has a unique generic point. Proof. Let Z be an irreducible closed subset of X: For any affine open subset U of X; we consider the intersection U \ Z: Either U \ Z is empty or U \ Z is nonempty. Suppose that U \ Z is nonempty. Then U \ Z is an irreducible closed subset of U: Write U = Spec A: Then U \ Z = V (I) for some ideal I: Since U \ Z is irreducible, I = p is a prime ideal. Then V (p) is the closure of p in U \ Z: Hence p is the generic point of U \ Z: Let ξ be the generic point of U \ Z: Similar we can show that U \ Z share the same generic point ξ for all affine open subset U of X: Using the fact that affine open subsets forms a base for the Zariski topology of X; ξ is the generic point of Z: (The detailed proof of these statements is the same as that of the above proposition.) 3 Proposition 1.6. Let X be a scheme. T.F.A.E. (1) X is irreducible. (2) There exists an affine open covering fUig so that Ui is irreducible and Ui \ Uj 6= ; for all i; j: (3) X is nonempty and every nonempty affine open U ⊂ X is irreducible. Proof. (1) implies (2). Since X is irreducible, it has a unique generic point ξ: Since X is a scheme, we can choose an affine open covering fUig for X so that Ui are all nonempty. Since Ui is nonempty, Ui contains the generic point ξ for all i: This implies that Ui \ Uj contains ξ and hence is nonempty. Moreover any subspace of irreducible topological space is also irreducible. Hence Ui is irreducible. Similarly, (1) implies (3). (2) implies (1). Let X = Z1 [ Z2 where Z1 and Z2 are closed subsets of X: For any Ui; either Ui \ Z1 = ; or Ui \ Z2 = ;: Hence either Ui ⊂ Z1 or Ui ⊂ Z2: We may assume that there exists Ui so that Ui ⊂ Z1: Since Ui is irreducible, Ui \ Z1 = Ui is an irreducible closed subset of Ui: Then Ui \ Z1 has a unique generic point ξ: For any Uj; we know that Ui \ Uj is nonempty by assumption and hence contains ξ: Then Ui \ Uj is densed in Uj; and contained in Z1 \ Uj: Notice that Z1 \ Uj is a closed set containing ξ; then Z1 \ Uj contains the closure of ξ in Uj: Since Ui \ Uj is dense in Uj; the closure of ξ in Uj is Uj: S Hence Uj ⊂ Z1 \ Uj ⊂ Uj: We conclude that Uj ⊂ Z1: Hence X = i Ui ⊂ Z1 which implies that X = Z1: Thus X is irreducible.
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